Calculus Examples

${\sec (2 x)}^{\cos (2 x)}$

Step 1

Use the properties of logarithms to simplify the differentiation.

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Step 1.1

Rewrite ${\sec (2 x)}^{\cos (2 x)}$ as $e^{\ln ({\sec (2 x)}^{\cos (2 x)})}$ .

$\frac{d}{d x} [e^{\ln ({\sec (2 x)}^{\cos (2 x)})}]$

Step 1.2

Expand $\ln ({\sec (2 x)}^{\cos (2 x)})$ by moving $\cos (2 x)$ outside the logarithm.

$\frac{d}{d x} [e^{\cos (2 x) \ln (\sec (2 x))}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = \cos (2 x) \ln (\sec (2 x))$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $\cos (2 x) \ln (\sec (2 x))$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [\cos (2 x) \ln (\sec (2 x))]$

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [\cos (2 x) \ln (\sec (2 x))]$

Step 2.3

Replace all occurrences of $u_{1}$ with $\cos (2 x) \ln (\sec (2 x))$ .

$e^{\cos (2 x) \ln (\sec (2 x))} \frac{d}{d x} [\cos (2 x) \ln (\sec (2 x))]$

Step 3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \cos (2 x)$ and $g (x) = \ln (\sec (2 x))$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos (2 x) \frac{d}{d x} [\ln (\sec (2 x))] + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = \sec (2 x)$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $\sec (2 x)$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos (2 x) (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [\sec (2 x)]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 4.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos (2 x) (\frac{1}{u_{2}} \frac{d}{d x} [\sec (2 x)]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 4.3

Replace all occurrences of $u_{2}$ with $\sec (2 x)$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos (2 x) (\frac{1}{\sec (2 x)} \frac{d}{d x} [\sec (2 x)]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 5

Rewrite $\sec (2 x)$ in terms of sines and cosines.

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos (2 x) (\frac{1}{\frac{1}{\cos (2 x)}} \frac{d}{d x} [\sec (2 x)]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 6

Multiply by the reciprocal of the fraction to divide by $\frac{1}{\cos (2 x)}$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos (2 x) (1 \cos (2 x) \frac{d}{d x} [\sec (2 x)]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 7

Multiply $\cos (2 x)$ by $1$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos (2 x) (\cos (2 x) \frac{d}{d x} [\sec (2 x)]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 8

Raise $\cos (2 x)$ to the power of $1$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos^{1} (2 x) \cos (2 x) \frac{d}{d x} [\sec (2 x)] + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 9

Raise $\cos (2 x)$ to the power of $1$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos^{1} (2 x) \cos^{1} (2 x) \frac{d}{d x} [\sec (2 x)] + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 10

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$e^{\cos (2 x) \ln (\sec (2 x))} ({\cos (2 x)}^{1 + 1} \frac{d}{d x} [\sec (2 x)] + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 11

Add $1$ and $1$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos^{2} (2 x) \frac{d}{d x} [\sec (2 x)] + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 12

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 2 x$ .

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Step 12.1

To apply the Chain Rule, set $u_{3}$ as $2 x$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos^{2} (2 x) (\frac{d}{d u_{3}} [\sec (u_{3})] \frac{d}{d x} [2 x]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 12.2

The derivative of $\sec (u_{3})$ with respect to $u_{3}$ is $\sec (u_{3}) \tan (u_{3})$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos^{2} (2 x) (\sec (u_{3}) \tan (u_{3}) \frac{d}{d x} [2 x]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 12.3

Replace all occurrences of $u_{3}$ with $2 x$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos^{2} (2 x) (\sec (2 x) \tan (2 x) \frac{d}{d x} [2 x]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 13

Differentiate.

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Step 13.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos^{2} (2 x) \sec (2 x) \tan (2 x) (2 \frac{d}{d x} [x]) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 13.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos^{2} (2 x) \sec (2 x) \tan (2 x) (2 \cdot 1) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 13.3

Simplify the expression.

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Step 13.3.1

Multiply $2$ by $1$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (\cos^{2} (2 x) \sec (2 x) \tan (2 x) \cdot 2 + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 13.3.2

Move $2$ to the left of $\cos^{2} (2 x) \sec (2 x) \tan (2 x)$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (2 \cdot (\cos^{2} (2 x) \sec (2 x) \tan (2 x)) + \ln (\sec (2 x)) \frac{d}{d x} [\cos (2 x)])$

Step 14

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 14.1

To apply the Chain Rule, set $u_{4}$ as $2 x$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (2 \cos^{2} (2 x) \sec (2 x) \tan (2 x) + \ln (\sec (2 x)) (\frac{d}{d u_{4}} [\cos (u_{4})] \frac{d}{d x} [2 x]))$

Step 14.2

The derivative of $\cos (u_{4})$ with respect to $u_{4}$ is $- \sin (u_{4})$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (2 \cos^{2} (2 x) \sec (2 x) \tan (2 x) + \ln (\sec (2 x)) (- \sin (u_{4}) \frac{d}{d x} [2 x]))$

Step 14.3

Replace all occurrences of $u_{4}$ with $2 x$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (2 \cos^{2} (2 x) \sec (2 x) \tan (2 x) + \ln (\sec (2 x)) (- \sin (2 x) \frac{d}{d x} [2 x]))$

Step 15

Differentiate.

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Step 15.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (2 \cos^{2} (2 x) \sec (2 x) \tan (2 x) + \ln (\sec (2 x)) (- \sin (2 x) (2 \frac{d}{d x} [x])))$

Step 15.2

Multiply $2$ by $- 1$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (2 \cos^{2} (2 x) \sec (2 x) \tan (2 x) + \ln (\sec (2 x)) (- 2 \sin (2 x) \frac{d}{d x} [x]))$

Step 15.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (2 \cos^{2} (2 x) \sec (2 x) \tan (2 x) + \ln (\sec (2 x)) (- 2 \sin (2 x) \cdot 1))$

Step 15.4

Multiply $- 2$ by $1$ .

$e^{\cos (2 x) \ln (\sec (2 x))} (2 \cos^{2} (2 x) \sec (2 x) \tan (2 x) + \ln (\sec (2 x)) (- 2 \sin (2 x)))$

Step 16

Simplify.

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Step 16.1

Apply the distributive property.

$e^{\cos (2 x) \ln (\sec (2 x))} (2 \cos^{2} (2 x) \sec (2 x) \tan (2 x)) + e^{\cos (2 x) \ln (\sec (2 x))} (\ln (\sec (2 x)) (- 2 \sin (2 x)))$

Step 16.2

Move $2$ to the left of $e^{\cos (2 x) \ln (\sec (2 x))}$ .

$2 e^{\cos (2 x) \ln (\sec (2 x))} \cos^{2} (2 x) \sec (2 x) \tan (2 x) + e^{\cos (2 x) \ln (\sec (2 x))} \ln (\sec (2 x)) (- 2 \sin (2 x))$

Step 16.3

Reorder terms.

$2 e^{\cos (2 x) \ln (\sec (2 x))} \cos^{2} (2 x) \sec (2 x) \tan (2 x) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4

Simplify each term.

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Step 16.4.1

Rewrite $\sec (2 x)$ in terms of sines and cosines.

$2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \cos^{2} (2 x) \sec (2 x) \tan (2 x) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.2

Rewrite $\sec (2 x)$ in terms of sines and cosines.

$2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \cos^{2} (2 x) \frac{1}{\cos (2 x)} \tan (2 x) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.3

Cancel the common factor of $\cos (2 x)$ .

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Step 16.4.3.1

Factor $\cos (2 x)$ out of $2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \cos^{2} (2 x)$ .

$\cos (2 x) (2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \cos (2 x)) \frac{1}{\cos (2 x)} \tan (2 x) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.3.2

Cancel the common factor.

$\cos (2 x) (2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \cos (2 x)) \frac{1}{\cos (2 x)} \tan (2 x) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.3.3

Rewrite the expression.

$2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \cos (2 x) \tan (2 x) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.4

Rewrite in terms of sines and cosines, then cancel the common factors.

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Step 16.4.4.1

Add parentheses.

$2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} (\cos (2 x) \tan (2 x)) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.4.2

Reorder $\cos (2 x)$ and $\tan (2 x)$ .

$2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} (\tan (2 x) \cos (2 x)) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.4.3

Rewrite $2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \cos (2 x) \tan (2 x)$ in terms of sines and cosines.

$2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} (\frac{\sin (2 x)}{\cos (2 x)} \cos (2 x)) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.4.4

Cancel the common factors.

$2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \sin (2 x) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.5

Rewrite $\sec (2 x)$ in terms of sines and cosines.

$2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \sin (2 x) - 2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \sin (2 x) \ln (\sec (2 x))$

Step 16.4.6

Rewrite $\sec (2 x)$ in terms of sines and cosines.

$2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \sin (2 x) - 2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \sin (2 x) \ln (\frac{1}{\cos (2 x)})$

Step 16.5

Simplify each term.

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Step 16.5.1

Convert from $\frac{1}{\cos (2 x)}$ to $\sec (2 x)$ .

$2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) - 2 e^{\cos (2 x) \ln (\frac{1}{\cos (2 x)})} \sin (2 x) \ln (\frac{1}{\cos (2 x)})$

Step 16.5.2

Convert from $\frac{1}{\cos (2 x)}$ to $\sec (2 x)$ .

$2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\frac{1}{\cos (2 x)})$

Step 16.5.3

Convert from $\frac{1}{\cos (2 x)}$ to $\sec (2 x)$ .

$2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) - 2 e^{\cos (2 x) \ln (\sec (2 x))} \sin (2 x) \ln (\sec (2 x))$